If all suspicions are correct, we can correctly describe all numbers integrally represented by this polynomial: positive or negative are unimportant, most prime factors are unimportant, all that matters is that every exponent of a prime factor $q = 3 u^2 + 2 u v + 4 v^2$ must be divisible by 3.

I should have done this last time: most of the class field part has already been done, by Hudson and Williams (1991), Theorem 1 and Table 1 on page 134. You get my version of the polynomial by negating their variable $x.$

joro asked about high powers being represented primitively. It turns out that the polynomial is not divisible by 8 unless $a,b,c$ are all even. This, despite the fact that 2 is represented. I believe this happens for all the (unrepresented) primes $q = 3 u^2 + 2 u v + 4 v^2$ as well, in the strongest manner: the polynomial is not divisible by $q$ itself unless $a,b,c$ are. I thought there might be trouble with the prime 11, but no. Anyway, here are some prime powers represented primitively, where $47 = 36 + 11$ and $53 = 9 + 44:$

Can $f$ represents large power with coprime $a,b,c$? In the bivariate case this is forbidden by an abc related conjecture.
–
joroApr 11 '13 at 13:02

@joro, I don't believe that will be a problem here, but I will do some checking. The main thing is that this behaves very much as the principal binary quadratic form of a discriminant, there is a rule for multiplication.
–
Will JagyApr 11 '13 at 19:33

Sure, there are large powers with primitive representations. This follows from the multiplicativity that Will J. already noted. For instance, $f(3356898, 3732782, 5764967) = 7^{23}$ (I started from $f(2,1,0)=7$ and used $7^{23}=1^{10}7^{23}$ and $f(2,-1,0)=1$ to reduce the resulting $(a,b,c)=(472709258936428, 396738620092614, 257006830281609)$ ).
–
Noam D. ElkiesApr 11 '13 at 20:32

@Noam, thanks. I've been checking, it appears the only primes to worry about are 2 and 11. I do think that the polynomial is not divisible by 8 unless all three variables are even ( a small finite check mod 8, not done yet). Perhaps something similar for 11, not sure yet.
–
Will JagyApr 11 '13 at 21:12

$z^{k-1} x + y^k$ is surjective for all $k$ so large powers are represented primitively in arbitrary large degree with $z=1$.
–
joroApr 12 '13 at 8:16

1 Answer
1

Your conjectures are correct. So was the "someone else at MSRI [who]
muttered something about norm forms" (mentioned in earlier edits
of the question), except for the part about laughing at you.

As you in effect note, $f(a,b,c)$ is the norm $N_{K/{\bf Q}}(a+bx+cx^2)$,
where $x$ is one of the roots of $x^3-x^2-x-1 = 0$
and $K$ is the cubic number field ${\bf Q}(x)$.
This field has discriminant $-44$,
and ${\bf Z}[x]$ is the full ring of integers $O_K$
(equivalently, the field discriminant of $K/{\bf Q}$
equals the polynomial discriminant of $x^3-x^2-x-1$;
to check this in gp, compute

poldisc(x^3-x^2-x-1)
nfdisc(x^3-x^2-x-1)

and observe that both return $-44$). Now for (A),
you already know that $x^3-x^2-x-1$ has at least one root modulo
any prime $q$ unless $q$ is represented by the nonprincipal quadratic
form $3u^2+2uv+4v^2$ of discriminant $-44$. (For other $q$:
there's a triple root for $q=2$, a double and a simple root for $q=11$,
three distinct roots for $q=u^2+11v^2$, and one simple root for odd
$q$ not congruent to a square $\bmod 11$.) Equivalently, $K$ has
an ideal of norm $q$ unless $q = 3u^2+2uv+4v^2$. But $O_K$ is
a principal ideal domain, so once there's an ideal of norm $q$
then it has a generator $a+bx+cx^2 \in O_K$, and then $q=f(a,b,c)$
(or $q=f(-a,-b,-c)$ if we chose $a+bx+cx^2$ of norm $-q$). The discriminant of $K$
is small enough that one can check unique factorization by hand using
the Minkowski bound; nowadays this exercise can also be done routinely
on the computer, e.g. in gp

[EDIT In fact this $K$ happens to be one of the handful of number fields
whose Minkowski bound is so tight that nothing needs to be checked!
The discriminant $\Delta_K = -44$ is small enough in absolute value that
the bound
$$
\frac4\pi \frac{3!}{3^3} \left|\Delta_K\right|^{1/2} = 1.8768\ldots
$$
is less than $2$, which means every ideal $I$ has a nonzero element
of norm $\pm \left|I\right|$ and is thus automatically principal. TIDE]

(B) Translating the factorization of $x^3-x^2-x-1 \bmod q$
into the factorization of the ideal $(q)$ in $O_K$, we see that
if $q = 3u^2+2uv+4v^2$ then $(q)$ remains prime in $O_K$,
and thus that $q \mid N_{K/{\bf Q}}(a+bx+cx^2)$ iff $q \mid a+bx+cx^2$.
For $q=2$ the ideal $(q)$ is the cube of $(1+x)$, so $8 \mid f(a,b,c)$
iff $a,b,c$ are all even. Any power of a prime $q$ other than
those of the form $3u^2+2uv+4v^2$ can be represented primitively by $f$,
even $q=11$ (for which $(q)$ factors as $(2+x)(3-2x)^2$). If we do not
care about primitivity then we can also represent all powers of $2$,
and all powers of $q^3$ for $q = 3u^2+2uv+4v^2$.

By multiplicativity this also proves the final conjecture:
the nonzero $n \in {\bf Z}$ that are represented by $f$
are precisely those whose $q$-valuation is a multiple of $3$
for all primes $q = 3u^2+2uv+4v^2$.